Proof that Integral over the arc vanishes as $R\rightarrow\infty$ in Inverse Mellin Transform of $\Gamma(s)$

Question

It´s a very well know result that $$e^{-x}=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(s)x^{-s}ds$$ In order to solve this integral we have to close the contour to the left and show that the integral over this path vanishes as $R\rightarrow\infty$. Unfortunately it´s not obvious to me how to proof it. All textbooks that I came across assume that the integral goes to zero without any proof.

What I have tried so far is the following: I assume that we are closing this contour with a semi circle to the left, and I get the following integral: Let $s=Re^{i\theta}$ and $ds=iRe^{i\theta}d\theta$, $$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \Gamma(Re^{i\theta})x^{-Re^{i\theta}}iRe^{i\theta}d\theta$$ Using Stirling´s approximation for the Gamma function we have $$\Gamma(Re^{i\theta})\sim\sqrt{2\pi} e^{-Re^{i\theta}}(Re^{i\theta})^{Re^{i\theta}-\frac{1}{2}}$$ and $$\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \Gamma(Re^{i\theta})x^{-Re^{i\theta}}iRe^{i\theta}d\theta\sim\ \sqrt{2\pi} iR \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{-Re^{i\theta}}(Re^{i\theta})^{Re^{i\theta}-\frac{1}{2}}x^{-Re^{i\theta}}e^{i\theta}d\theta$$

I took the modulus of the last integral and arrived in the following expression:

$$\sim \sqrt{2 \pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{Rcos(\theta)(log(R)-log(x)-1)}e^{\frac{1}{2}log(R)}e^{-R \theta sin(\theta)}d\theta$$

But I don´t know how to evaluate the $\lim$ as $R \to \infty$

I really appreciate if someone could show this for me. Thank you in advance.

Answer 1

I finally got the answer!

The integrand $$\sqrt{2 \pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{Rcos(\theta)(log(R)-log(x)-1)}e^{\frac{1}{2}log(R)}e^{-R \theta sin(\theta)}d\theta$$

could be rewritten as follows:

$$\sim \sqrt{\frac{2\pi}{R}} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{Rcos(\theta)(log(\frac{R}{ex})}e^{-R \theta sin(\theta)}d\theta$$

Now, if we choose $R>ex$ and note that for $\frac{\pi}{2}<\theta<\frac{3\pi}{2}$, $\cos(\theta)<0$ and $\theta\sin(\theta)>0$, in the limit $\lim{R}\rightarrow\infty$ the integrand goes to zero.

Answer 2

Apply Jordan's lemma:

http://mathworld.wolfram.com/JordansLemma.html

We want to bound $$I_R = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\Gamma(s)x^{-s}ds, \quad \textrm{ where } s=Re^{i\theta}.$$

Let $p=\displaystyle \frac{s}{i}$, then

$$I_R = i\int_{0}^{\pi}\Gamma(ip)x^{ip}dp, \quad \textrm{ where } p=Re^{i\theta}.$$

$$I_R \sim \sqrt{2\pi} \int_{0}^\pi (ip)^{ip-\frac{1}{2} }\exp\left[1-ip\right]\exp[i (\ln x) p] \, dp = \int_0^\pi f(p) \exp[i(\ln x-1) p] \, dp$$

but, for $x>e, \ln x >1$ and $\lim_{R\rightarrow \infty} f(R e^{i\theta}) \rightarrow 0$, so by Jordan's lemma, $$I_R \rightarrow 0.$$

For $0<x<e$, we can make a change of variables to scale the original integral to guarantee that we can apply Jordan's lemma directly, so there too, the integral along the semicircular contour goes to zero.

In fact, the inverse Mellin transform is defined for $x \in (0,\infty)$.

Another point: You have taken a contour to be a semicircle, but $\gamma$ is required to be greater than zero and we should show that the integrals along the segment connecting $iR$ to $\gamma+iR$ and connecting $-iR$ to $\gamma-iR$ also go to zero. This shouldn't be difficult because most textbook treatments (see for example the link @Gary gave in the comments) use a rectangular contour that contains these.

Answer 3

• $\Gamma(\sigma+it)$ is the Fourier transform of $e^{-e^x+\sigma x}$ which is $L^1$ with $L^1$ second derivative, thus for $\sigma\in [1/2,3/2]$, $|\Gamma(\sigma+it)|\le \frac{A}{1+|t|^2}$ where $A$ is the $\sup$ of the mentioned $L^1$ norms.

• $\Gamma(s+1)=s\Gamma(s)$ gives $$|\Gamma(1/2-N+it)|\le 2\frac{|\Gamma(1/2+it)|}{|t|^N}$$

  Both facts imply that $$\int_{[1/2-iN,1/2+iN]\cup [1/2-N\pm iN,1/2-N\pm iN]} |\Gamma(s)x^{-s}|d|s|\to 0 \text{ as } N\to \infty$$

• Since $\Gamma(s)$ is analytic away from the poles at negative integers then Cauchy's integral theorem implies that
  $$\int_{|s| = N-1/2, \Re(s)<1/2} |\Gamma(s)x^{-s}|d|s|\to 0 \text{ as } N\to \infty$$

• And hence

$$\frac1{2i\pi}\int_{1/2-i\infty}^{1/2+i\infty}\Gamma(s)x^{-s}ds = \sum_{k\ge 0} Res(\Gamma(s)x^{-s},k) = e^{-x}$$